Question

If $\int x^5(1+x^3{)}^{2/3}dx=A(1+x^3{)}^{8/3}+B(1+x^3{)}^{5/3}+c$, then

• A. $A=\frac{1}{4},B=\frac{1}{5}$
• B. $A=\frac{1}{8},B=-\frac{1}{5}$
• C. $A=-\frac{1}{8},B=\frac{1}{5}$
• D. None of these

Answer 1

Answer: (A)

$A=\frac{1}{4},B=\frac{1}{5}$

$\int x^5{(1+x^3)}^{2/3}dx$

Let $x^3=t$ $\Rightarrow {3x}^{2}dx=dt$

$=\int x^3{x}^2{(1+x^3)}^{2/3}dx$

$=\int t{(1+t)}^{2/3}\frac{dt}{3}$

$=\frac{1}{3}[t\int {(1+t)}^{2/3}dt-\int \frac{dt}{dt}(\int {(1+t)}^{2/3}dt)dt]$

$=\frac{1}{3}[t\frac{{(1+t)}^{5/3}}{5/3}-\int \frac{{(1+t)}^{5/3}}{5/3}dt]$

$=\frac{1}{3}[\frac{3t}{5}{(1+t)}^{5/3}-\frac{3}{5}\frac{{(1+t)}^{8/3}}{8/3}]$

$=\frac{{t(1+t)}^{5/3}}{5}-\frac{3}{40}{(1+t)}^{8/3}$

So, $A=-\frac{3}{40}$

$=\frac{x^3{(1+x^3)}^{5/3}}{5}-\frac{3}{40}{(1+x^3)}^{8/3}$

$B=\frac{1}{5}$ (Answer)